Symfony 4中的Guard Authenticator

Question

我正在Symfony 4中创建一个简单的登录身份验证系统，并使用安全组件Guard。我的FormLoginAuthenticator如下：

<?php
namespace App\Security;

use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\RedirectResponse;
use Symfony\Component\Routing\RouterInterface;
use Symfony\Component\Security\Core\Encoder\UserPasswordEncoderInterface;
use Symfony\Component\Security\Core\Exception\BadCredentialsException;
use Symfony\Component\Security\Core\Exception\AuthenticationException;
use Symfony\Component\Security\Core\User\UserInterface;
use Symfony\Component\Security\Core\User\UserProviderInterface;
use Symfony\Component\Security\Core\Authentication\Token\TokenInterface;
use Symfony\Component\Security\Guard\Authenticator\AbstractFormLoginAuthenticator;
use Symfony\Component\Security\Core\Security;

class FormLoginAuthenticator extends AbstractFormLoginAuthenticator
{
    private $router;
    private $encoder;

    public function __construct(RouterInterface $router, UserPasswordEncoderInterface $encoder)
    {
        $this->router = $router;
        $this->encoder = $encoder;
    }

    public function getCredentials(Request $request)
    {
        if ($request->getPathInfo() != '/login_check') {
          return;
        }

        $email = $request->request->get('_email');
        $request->getSession()->set(Security::LAST_USERNAME, $email);
        $password = $request->request->get('_password');

        return [
            'email' => $email,
            'password' => $password,
        ];
    }

    public function getUser($credentials, UserProviderInterface $userProvider)
    {
        $email = $credentials['email'];

        return $userProvider->loadUserByUsername($email);
    }

    public function checkCredentials($credentials, UserInterface $user)
    {
        $plainPassword = $credentials['password'];
        if ($this->encoder->isPasswordValid($user, $plainPassword)) {
            return true;
        }

        throw new BadCredentialsException();
    }

    public function onAuthenticationSuccess(Request $request, TokenInterface $token, $providerKey)
    {
        $url = $this->router->generate('welcome');

        return new RedirectResponse($url);
    }

    public function onAuthenticationFailure(Request $request, AuthenticationException $exception)
    {
       $request->getSession()->set(Security::AUTHENTICATION_ERROR, $exception);

       $url = $this->router->generate('login');

       return new RedirectResponse($url);
    }

    protected function getLoginUrl()
    {
        return $this->router->generate('login');
    }

    protected function getDefaultSuccessRedirectUrl()
    {
        return $this->router->generate('welcome');
    }

    public function supportsRememberMe()
    {
        return false;
    }
}

但它显示以下错误：

(1/1) FatalErrorException错误: App\Security\FormLoginAuthenticator类包含1个抽象方法，因此必须声明为抽象方法或实现其余方法FormLoginAuthenticator

你能告诉我这个错误是由这个类的哪个抽象方法引起的吗？

Answer 1

Symfony\Component\Security\Guard\Authenticator\AbstractFormLoginAuthenticator的子类必须实现以下抽象方法：

/**
 * Return the URL to the login page.
 *
 * @return string
 */
protected function getLoginUrl()

/**
 * Does the authenticator support the given Request?
 *
 * If this returns false, the authenticator will be skipped.
 *
 * @param Request $request
 *
 * @return bool
 */
public function supports(Request $request)

/**
 * Get the authentication credentials from the request and return them
 * as any type (e.g. an associate array).
 *
 * Whatever value you return here will be passed to getUser() and checkCredentials()
 *
 * For example, for a form login, you might:
 *
 *      return array(
 *          'username' => $request->request->get('_username'),
 *          'password' => $request->request->get('_password'),
 *      );
 *
 * Or for an API token that's on a header, you might use:
 *
 *      return array('api_key' => $request->headers->get('X-API-TOKEN'));
 *
 * @param Request $request
 *
 * @return mixed Any non-null value
 *
 * @throws \UnexpectedValueException If null is returned
 */
public function getCredentials(Request $request)

/**
 * Return a UserInterface object based on the credentials.
 *
 * The *credentials* are the return value from getCredentials()
 *
 * You may throw an AuthenticationException if you wish. If you return
 * null, then a UsernameNotFoundException is thrown for you.
 *
 * @param mixed $credentials
 * @param UserProviderInterface $userProvider
 *
 * @throws AuthenticationException
 *
 * @return UserInterface|null
 */
public function getUser($credentials, UserProviderInterface $userProvider)

/**
 * Returns true if the credentials are valid.
 *
 * If any value other than true is returned, authentication will
 * fail. You may also throw an AuthenticationException if you wish
 * to cause authentication to fail.
 *
 * The *credentials* are the return value from getCredentials()
 *
 * @param mixed $credentials
 * @param UserInterface $user
 *
 * @return bool
 *
 * @throws AuthenticationException
 */
public function checkCredentials($credentials, UserInterface $user)

/**
 * Called when authentication executed and was successful!
 *
 * This should return the Response sent back to the user, like a
 * RedirectResponse to the last page they visited.
 *
 * If you return null, the current request will continue, and the user
 * will be authenticated. This makes sense, for example, with an API.
 *
 * @param Request $request
 * @param TokenInterface $token
 * @param string $providerKey The provider (i.e. firewall) key
 *
 * @return Response|null
 */
public function onAuthenticationSuccess(Request $request, TokenInterface $token, $providerKey)

正如您在错误中看到的，它是由于缺少方法supports而发生的

Answer 2

在FormLoginAuthenticator类中编写：

public function supports(Request $request)
{
    return 'app_login' === $request->attributes->get('_route')
    && $request->isMethod('POST');
}

'app_login‘是名称控制器登录